Geotechnical project specifications require that the soil's optimum moisture content of 24% be maintained during the roadbed's construction.   The field moisture test finds that five-pounds of soil has a water content of 11%.  The amount of water that must be added during the day's planned production of 1,250-tons to achieve the optimum moisture content is most nearly:

Solution:  The total mass of the moist soil is equal to the dry soil and water content.  The mass of water is equal to 11% of the dry soil.  This relationship is represented in the following equation:

M_total     = M_soil + M_water

M_total     = M_soil + 0.11 M_soil

M_total     = 1.11 M_soil

Solve for the mass of the dry soil and water:

M_soil =  M_total/1.11  =   5-lb/1.11  =    4.50-lb

M_water   = 0.11M_soil

 = (0.11) (4.50-lb) = .50-lb  

To raise the water content from 11% to 24%, the earthwork contractor must add 13% by mass of water.

 ΔM_water   = (ΔW_required) (M_soil)

    = (0.13) (4.50-lb) 

    = 0.585-lbm of water per 5-lbs of soil

OR  = 0.117-lb of water / lb of soil  

Convert the results of the required additional water and apply it to the day's planned production:

 1,250-ton x 2,000-lb/ton = 2,500,000-lb of soil

2,500,000-lb of soil x 0.117-lbm of water/lb of soil = 292,500-lbm of water

292,500-lb of water ÷ 62.4-lb of water/ft³ = 4,688-ft³

4,688-ft³ x 7.48gal/ft³ = 35,062.5-gallons